Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -\dfrac{3}{2}x + 2}\enspace$ and passes through the point ${(4, 3)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Solution: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-\dfrac{3}{2}}$ , and its negative reciprocal is ${\dfrac{2}{3}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = \dfrac{2}{3}x + b}\enspace$ We can plug our point, $(4, 3)$ , into this equation to solve for ${b}$ , the y-intercept. $3 = {\dfrac{2}{3}}(4) + {b}$ $3 = \dfrac{8}{3} + {b}$ $3 - \dfrac{8}{3} = {b} = \dfrac{1}{3}$ The equation of the perpendicular line is $\enspace {y = \dfrac{2}{3}x + \dfrac{1}{3}}\enspace$. ${m = \dfrac{2}{3}, \enspace b = \dfrac{1}{3}}$